Page 264 - Calculus Demystified

CHAPTER 8
Applications of the Integral
4√2 251

Fig. 8.41

h = 16
4√2
_
h 16

h = 24
Fig. 8.42

According to our analysis, the total force on the upper half of the window is thus
20

44880
h · 50 · (2h − 32)dh = lbs.
16 3
For the lower half of the window, we examine the isosceles right triangle in
Fig. 8.43. It has base 24 − h. Therefore, for h ranging from 20 to 24, we have

w(h) = 2(24 − h) = 48 − 2h.
According to our analysis, the total force on the lower half of the window is
24 51200
h · 50 · (48 − 2h) dh = lbs.
20 3
The total force on the entire window is thus
44880 51200 96080
+ = lbs.
3 3 3
You Try It: Atank of water has flat sides. In one side, with center 4 feet below the
surface of the water, is a circular window of radius 1 foot. What is the total force
on the window?